Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}7x-3y &= -5 \\ 8x-3y &= -1\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = 3y-1$ Divide both sides by $8$ to isolate $x$ $x = {\dfrac{3}{8}y - \dfrac{1}{8}}$ Substitute this expression for $x$ in the first equation. $7({\dfrac{3}{8}y - \dfrac{1}{8}}) - 3y = -5$ $\dfrac{21}{8}y - \dfrac{7}{8} - 3y = -5$ Simplify by combining terms, then solve for $y$ $-\dfrac{3}{8}y - \dfrac{7}{8} = -5$ $-\dfrac{3}{8}y = -\dfrac{33}{8}$ $y = 11$ Substitute $11$ for $y$ in the top equation. $7x-3( 11) = -5$ $7x-33 = -5$ $7x = 28$ $x = 4$ The solution is $\enspace x = 4, \enspace y = 11$.